Date modified: Fri 2024-03-22 . 02:15 AM
Date created: Fri 2024-03-22 . 01:38 AM
# 23a Bijective function if and only if invertible function. Let us prove the characterization > A function $f:A\to B$ is bijective if and only if $f$ is invertible. Before we start, we recall the following definitions: Given a function $f:A \to B$.... - $f$ is bijective if it is both surjective and injective. - $f$ is surjective if $\text{range}(f)=\text{codom}(f)=B$. - $f$ is injective if $f(x)=f(y)$ implies $x=y$, for all $x,y \in A$. - $f$ is invertible if there exists a function $g:B\to A$ such that $g\circ f = \text{id}_{A}$ and $f\circ g=\text{id}_{B}$. We often denote such $g$ as $f^{-1}$. $\blacktriangleright$ Proof. $(\implies)$ Let us assume $f:A\to B$ is is bijective. We show $f$ is invertible by constructing an inverse function $g:B\to A$. Since $f$ is surjective, for each $b\in B$, there exists some $a\in A$ such that $f(a) =b$, in particular the preimage $f^{\leftarrow}(b)$ is not empty. And we claim this preimage is just a singleton set $\{a\}$. Indeed, if we have $a,a'$ such that $f(a)=b$ and $f(a')=b$, then $f(a)=f(a')$. But as $f$ is injective, we have $a=a'$. Conclusion: For each $b\in B$ there exists precisely one $a\in A$ where $f(a)=b$. This allows to define the function $g:B\to A$ as follows: For each $b\in B$, let $g(b)$ be this unique $a\in A$ such that $f(a)=b$. This $g$ is well-defined. Now we show this constructed $g$ is indeed the inverse of $f$. By its construction, for each $b\in B$, $f(g(b))=b$, so $f\circ g=\text{id}_{B}$. We now also claim $g(f(x))=x$ for each $x\in A$. Indeed, say $f(x)=y$, then $g(y)$ is the unique point in $A$ such that $f(g(y))=y$. So we see that $f(x)=y=f(g(y))$, by injectivity of $f$ we have $x=g(y)$. Hence $x=g(f(x))$ for each $x\in A$, namely $g\circ f =\text{id}_{A}$. This shows $g$ is indeed an inverse of $f$, and $f$ is invertible. $(\impliedby)$ Now, suppose $f$ is invertible. Then there exists a function $g:B\to A$ such that $g\circ f=\text{id}_{A}$ and $f\circ g=\text{id}_{B}$. Let us show $f$ is injective. Suppose $f(x)=f(y)$, then apply $g$ to both sides we get $g(f(x))=g(f(y))$, which implies $x=y$. Hence $f$ is injective. Now to show $f$ is surjective, take $y\in B$, and we need to show there exists some $x\in A$ where $f(x)=b$. Indeed, consider the point $g(y)\in A$, we have $f(g(y))=y$. So $f$ is surjective. Hence $f$ is bijective as $f$ is both surjective and injective. $\blacksquare$